1 /* @(#)e_jn.c 5.1 93/09/24 */
3 * ====================================================
4 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
6 * Developed at SunPro, a Sun Microsystems, Inc. business.
7 * Permission to use, copy, modify, and distribute this
8 * software is freely granted, provided that this notice
10 * ====================================================
13 #if defined(LIBM_SCCS) && !defined(lint)
14 static char rcsid
[] = "$NetBSD: e_jn.c,v 1.9 1995/05/10 20:45:34 jtc Exp $";
18 * __ieee754_jn(n, x), __ieee754_yn(n, x)
19 * floating point Bessel's function of the 1st and 2nd kind
23 * y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal;
24 * y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
25 * Note 2. About jn(n,x), yn(n,x)
26 * For n=0, j0(x) is called,
27 * for n=1, j1(x) is called,
28 * for n<x, forward recursion us used starting
29 * from values of j0(x) and j1(x).
30 * for n>x, a continued fraction approximation to
31 * j(n,x)/j(n-1,x) is evaluated and then backward
32 * recursion is used starting from a supposed value
33 * for j(n,x). The resulting value of j(0,x) is
34 * compared with the actual value to correct the
35 * supposed value of j(n,x).
37 * yn(n,x) is similar in all respects, except
38 * that forward recursion is used for all
44 #include "math_private.h"
52 invsqrtpi
= 5.64189583547756279280e-01, /* 0x3FE20DD7, 0x50429B6D */
53 two
= 2.00000000000000000000e+00, /* 0x40000000, 0x00000000 */
54 one
= 1.00000000000000000000e+00; /* 0x3FF00000, 0x00000000 */
57 static const double zero
= 0.00000000000000000000e+00;
59 static double zero
= 0.00000000000000000000e+00;
63 double attribute_hidden
__ieee754_jn(int n
, double x
)
65 double attribute_hidden
__ieee754_jn(n
,x
)
69 int32_t i
,hx
,ix
,lx
, sgn
;
70 double a
, b
, temp
=0, di
;
73 /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
74 * Thus, J(-n,x) = J(n,-x)
76 EXTRACT_WORDS(hx
,lx
,x
);
78 /* if J(n,NaN) is NaN */
79 if((ix
|((u_int32_t
)(lx
|-lx
))>>31)>0x7ff00000) return x
+x
;
85 if(n
==0) return(__ieee754_j0(x
));
86 if(n
==1) return(__ieee754_j1(x
));
87 sgn
= (n
&1)&(hx
>>31); /* even n -- 0, odd n -- sign(x) */
89 if((ix
|lx
)==0||ix
>=0x7ff00000) /* if x is 0 or inf */
91 else if((double)n
<=x
) {
92 /* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
93 if(ix
>=0x52D00000) { /* x > 2**302 */
95 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
96 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
97 * Let s=sin(x), c=cos(x),
98 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
100 * n sin(xn)*sqt2 cos(xn)*sqt2
101 * ----------------------------------
108 case 0: temp
= cos(x
)+sin(x
); break;
109 case 1: temp
= -cos(x
)+sin(x
); break;
110 case 2: temp
= -cos(x
)-sin(x
); break;
111 case 3: temp
= cos(x
)-sin(x
); break;
113 b
= invsqrtpi
*temp
/sqrt(x
);
119 b
= b
*((double)(i
+i
)/x
) - a
; /* avoid underflow */
124 if(ix
<0x3e100000) { /* x < 2**-29 */
125 /* x is tiny, return the first Taylor expansion of J(n,x)
126 * J(n,x) = 1/n!*(x/2)^n - ...
128 if(n
>33) /* underflow */
131 temp
= x
*0.5; b
= temp
;
132 for (a
=one
,i
=2;i
<=n
;i
++) {
133 a
*= (double)i
; /* a = n! */
134 b
*= temp
; /* b = (x/2)^n */
139 /* use backward recurrence */
141 * J(n,x)/J(n-1,x) = ---- ------ ------ .....
142 * 2n - 2(n+1) - 2(n+2)
145 * (for large x) = ---- ------ ------ .....
147 * -- - ------ - ------ -
150 * Let w = 2n/x and h=2/x, then the above quotient
151 * is equal to the continued fraction:
153 * = -----------------------
155 * w - -----------------
160 * To determine how many terms needed, let
161 * Q(0) = w, Q(1) = w(w+h) - 1,
162 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
163 * When Q(k) > 1e4 good for single
164 * When Q(k) > 1e9 good for double
165 * When Q(k) > 1e17 good for quadruple
169 double q0
,q1
,h
,tmp
; int32_t k
,m
;
170 w
= (n
+n
)/(double)x
; h
= 2.0/(double)x
;
171 q0
= w
; z
= w
+h
; q1
= w
*z
- 1.0; k
=1;
179 for(t
=zero
, i
= 2*(n
+k
); i
>=m
; i
-= 2) t
= one
/(i
/x
-t
);
182 /* estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
183 * Hence, if n*(log(2n/x)) > ...
184 * single 8.8722839355e+01
185 * double 7.09782712893383973096e+02
186 * long double 1.1356523406294143949491931077970765006170e+04
187 * then recurrent value may overflow and the result is
188 * likely underflow to zero
192 tmp
= tmp
*__ieee754_log(fabs(v
*tmp
));
193 if(tmp
<7.09782712893383973096e+02) {
194 for(i
=n
-1,di
=(double)(i
+i
);i
>0;i
--){
202 for(i
=n
-1,di
=(double)(i
+i
);i
>0;i
--){
208 /* scale b to avoid spurious overflow */
216 b
= (t
*__ieee754_j0(x
)/b
);
219 if(sgn
==1) return -b
; else return b
;
223 double attribute_hidden
__ieee754_yn(int n
, double x
)
225 double attribute_hidden
__ieee754_yn(n
,x
)
233 EXTRACT_WORDS(hx
,lx
,x
);
235 /* if Y(n,NaN) is NaN */
236 if((ix
|((u_int32_t
)(lx
|-lx
))>>31)>0x7ff00000) return x
+x
;
237 if((ix
|lx
)==0) return -one
/zero
;
238 if(hx
<0) return zero
/zero
;
242 sign
= 1 - ((n
&1)<<1);
244 if(n
==0) return(__ieee754_y0(x
));
245 if(n
==1) return(sign
*__ieee754_y1(x
));
246 if(ix
==0x7ff00000) return zero
;
247 if(ix
>=0x52D00000) { /* x > 2**302 */
249 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
250 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
251 * Let s=sin(x), c=cos(x),
252 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
254 * n sin(xn)*sqt2 cos(xn)*sqt2
255 * ----------------------------------
262 case 0: temp
= sin(x
)-cos(x
); break;
263 case 1: temp
= -sin(x
)-cos(x
); break;
264 case 2: temp
= -sin(x
)+cos(x
); break;
265 case 3: temp
= sin(x
)+cos(x
); break;
267 b
= invsqrtpi
*temp
/sqrt(x
);
272 /* quit if b is -inf */
273 GET_HIGH_WORD(high
,b
);
274 for(i
=1;i
<n
&&high
!=0xfff00000;i
++){
276 b
= ((double)(i
+i
)/x
)*b
- a
;
277 GET_HIGH_WORD(high
,b
);
281 if(sign
>0) return b
; else return -b
;